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4. ) For each of the statements below (which all happen to be false), look at the "proofs" that have been provided and explain in one or two sentences what error has been mnade by the author. (Note: We are not looking for picky details about writing style here, but actual major problems with the proofs.) (a) Statement: For all odd integers m and n, mtn is an integer. "Proof": Let m = 3 and let n = 5. By definition of odd, 3 is odd because 3 = 2(1) + 1, and 1 is an integer. Similarly, 5 is odd because 5 = 2(2) + 1, and 2 is an integer. We wish to prove that is an integer. 3+5 * (by substitution) 32 (by algebra) (by algebra) = 7 (by algebra) Since 7 is an integer, we have shown what we were required to show. O (b) Statement: For all integers m, if 6m – 10 is even, then m is odd. "Proof": Let m be any odd integer. WWe wish to prove that 6m- 10 is even. In other words, we wish to show that 6m – 10 = 2k for some integer k. Since m is odd, we know that m = 2p + 1 for some integer p (by definition of odd). Now, 6m - 10 = 6(2p + 1) – 10 = 12p + 6 – 10 = 12p – 4 = 2(6p – 2) (by substitution) (by algebra) (by algebra) (by algebra) We know that 6p - 2 is an integer, since it is the product and difference of integers (6, p, 2). 10 = 2k for an integer k. By definition of even, 6m - 10 Therefore, we have shown that 6m is even. O

4. ) For each of the statements below (which all happen to be false), look at the "proofs" that have been provided and explain in one or two sentences what error has been mnade by the author. (Note: We are not looking for picky details about writing style here, but actual major problems with the proofs.) (a) Statement: For all odd integers m and n, mtn is an integer. "Proof": Let m = 3 and let n = 5. By definition of odd, 3 is odd because 3 = 2(1) + 1, and 1 is an integer. Similarly, 5 is odd because 5 = 2(2) + 1, and 2 is an integer. We wish to prove that is an integer. 3+5 * (by substitution) 32 (by algebra) (by algebra) = 7 (by algebra) Since 7 is an integer, we have shown what we were required to show. O (b) Statement: For all integers m, if 6m – 10 is even, then m is odd. "Proof": Let m be any odd integer. WWe wish to prove that 6m- 10 is even. In other words, we wish to show that 6m – 10 = 2k for some integer k. Since m is odd, we know that m = 2p + 1 for some integer p (by definition of odd). Now, 6m - 10 = 6(2p + 1) – 10 = 12p + 6 – 10 = 12p – 4 = 2(6p – 2) (by substitution) (by algebra) (by algebra) (by algebra) We know that 6p - 2 is an integer, since it is the product and difference of integers (6, p, 2). 10 = 2k for an integer k. By definition of even, 6m - 10 Therefore, we have shown that 6m is even. O

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.2: Mathematical Induction
Problem 49E: Show that if the statement is assumed to be true for , then it can be proved to be true for . Is...
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4. -) For each of the statements below (which all happen to be false), look at the "proofs" that have been provided and explain in one or two sentences what error has been made by the author. (Note: We are not looking for picky details about writing style here, but actual major problems with the proofs.) (a) Statement: For all odd integers m and n, mtn is an integer. "Proof": Let m = 3 and let n = 5. By definition of odd, 3 is odd because 3 = 2(1) + 1, and 1 is an integer. Similarly, 5 is odd because 5 = 2(2) + 1, and 2 is an integer. We wish to prove that ": is an integer. min = 315 (by substitution) 2 (by algebra) (by algebra) = 7 (by algebra) Since 7 is an integer, we have shown what we were required to show. O (b) Statement: For all integers m, if 6m – 10 is even, then m is odd. "Proof": Let m be any odd integer. We wish to prove that 6m - 10 is even. In other words, we wish to show that 6m – 10 = 2k for some integer k. Since m is odd, we know that m = 2p + 1 for some integer p (by definition of odd). Now, 6m- 10 = 6(2p + 1) – 10 = 12p + 6 – 10 12р — 4 3D 2(бр — 2) (by substitution) (by algebra) (by algebra) (by algebra) We know that 6p – 2 is an integer, since it is the product and difference of integers (6, p, 2). Therefore, we have shown that 6m - 10 = 2k for an integer k. By definition of even, 6m – 10 is even. O
Transcribed Image Text:4. -) For each of the statements below (which all happen to be false), look at the "proofs" that have been provided and explain in one or two sentences what error has been made by the author. (Note: We are not looking for picky details about writing style here, but actual major problems with the proofs.) (a) Statement: For all odd integers m and n, mtn is an integer. "Proof": Let m = 3 and let n = 5. By definition of odd, 3 is odd because 3 = 2(1) + 1, and 1 is an integer. Similarly, 5 is odd because 5 = 2(2) + 1, and 2 is an integer. We wish to prove that ": is an integer. min = 315 (by substitution) 2 (by algebra) (by algebra) = 7 (by algebra) Since 7 is an integer, we have shown what we were required to show. O (b) Statement: For all integers m, if 6m – 10 is even, then m is odd. "Proof": Let m be any odd integer. We wish to prove that 6m - 10 is even. In other words, we wish to show that 6m – 10 = 2k for some integer k. Since m is odd, we know that m = 2p + 1 for some integer p (by definition of odd). Now, 6m- 10 = 6(2p + 1) – 10 = 12p + 6 – 10 12р — 4 3D 2(бр — 2) (by substitution) (by algebra) (by algebra) (by algebra) We know that 6p – 2 is an integer, since it is the product and difference of integers (6, p, 2). Therefore, we have shown that 6m - 10 = 2k for an integer k. By definition of even, 6m – 10 is even. O
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